1.) The weight that will cause a wire of diameter d to stretch a given distance, for a fixed length of wire, is:
a) independent of d b) independent of what the length of the wire is. c) proportional to d d) proportional to d$\displaystyle ^2$

2.) Which one of the following would be expected to have the smallest bulk modulous?
a) solid iron b) helium vapor c) liquid mercury d) solid uranium e) liquid water

3.) The lowest tone to resonate in a closed pipe of length L is 200 Hz. Which of the following frequencies will not resonate in that pipe?
a) 600 Hz b) 400 Hz c) 1000 Hz d) 1400 Hz

4.) A trombone or trumpet can be considered a(n): a) closed organ pipe b) open organ pipe

Aug 30th 2007, 11:24 PM

John 5

*slight bump*

Aug 31st 2007, 01:51 PM

topsquark

Quote:

Originally Posted by John 5

*slight bump*

You may have noticed that "bumps" don't work very well here. In fact, they irritate a few members. Either your question will be answered or it won't.

I can help you with the last two:

Quote:

Originally Posted by John 5

3.) The lowest tone to resonate in a closed pipe of length L is 200 Hz. Which of the following frequencies will not resonate in that pipe?
a) 600 Hz b) 400 Hz c) 1000 Hz d) 1400 Hz

Something is wrong here. Read on.

Since both ends are closed the ends of the pipes must be nodes. That means the lowest tone (ie frequency) that will fit in the pipe is a waveform of 1/2 a wavelength. (There is a node at either end and an anti-node half-way down the pipe.) Thus the wavelength is 2L, and since it is the lowest frequency, 200 Hz is the fundamental frequency. Note that the speed of this wave is $\displaystyle v = f \lambda = (200)(2L) = 400L$. This is the wave speed no matter what the frequency is.

Now, all standing waves in this tube will have a node at each end. So the next lowest tone will have a wavelength of L (one whole wavelength fits the tube) so the frequency will be
$\displaystyle f = \frac{v}{\lambda} = \frac{400L}{L} = 400~Hz$

The next lowest frequency standing wave has 3/2 of a wavelength in the tube, so the wavelength is 2/3 of L:
$\displaystyle f = \frac{v}{\lambda} = \frac{400L}{\frac{2}{3}L} = 600~Hz$

etc.

So the general standing wave will have a wavelength of $\displaystyle \frac{2}{n}L$ where n is a positive integer. Thus the general frequency for this wave will be:
$\displaystyle f_n = \frac{v}{\lambda _n} = \frac{400L}{\frac{2}{n}L} = 200n~Hz$